给定一个二进制字符串S ,任务是为 DFA Machine 编写一个程序,该程序接受一个奇数为0和1的字符串。
例子:
Input: S = “010011”
Output: Accepted
Explanation:
The given string S contains odd number of zeros and ones.
Input: S = “00000”
Output: Not Accepted
Explanation:
The given string S doesn’t contains odd number of zeros and ones.
方法:下面是针对给定问题设计的 DFA 机器。为 DFA 状态构建一个转换表并分析每个状态之间的转换。以下是步骤:
- 有 4 个状态q 0 , q 1 , q 2 , q 3其中q 0是初始状态, q 3是最终状态。
- 上述DFA的转换表如下:
Current state | Final state | |
0 | 1 | |
q0 | q1 | q2 |
q1 | q0 | q3 |
q2 | q3 | q0 |
q3 | q2 | q1 |
- 通过此表,了解 DFA 中的转换。
- 如果在读取整个字符串后达到最终状态( q 3 ),则该字符串被接受,否则不被接受。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Function to check whether the given
// string is accepted by DFA or not
void checkValidDFA(string s)
{
// Stores initial state of DFA
int initial_state = 0;
// Stores final state of DFA
int final_state;
// Stores previous state of DFA
int previous_state = 0;
// Iterate through the string
for (int i = 0; i < s.length(); i++) {
// Checking for all combinations
if ((s[i] == '0'
&& previous_state == 0)
|| (s[i] == '1'
&& previous_state == 3)) {
final_state = 1;
}
else if ((s[i] == '0'
&& previous_state == 3)
|| (s[i] == '1'
&& previous_state == 0)) {
final_state = 2;
}
else if ((s[i] == '0'
&& previous_state == 1)
|| (s[i] == '1'
&& previous_state == 2)) {
final_state = 0;
}
else if ((s[i] == '0'
&& previous_state == 2)
|| (s[i] == '1'
&& previous_state == 1)) {
final_state = 3;
}
// Update the previous_state
previous_state = final_state;
}
// If final state is reached
if (final_state == 3) {
cout << "Accepted" << endl;
}
// Otherwise
else {
cout << "Not Accepted" << endl;
}
}
// Driver Code
int main()
{
// Given string
string s = "010011";
// Function Call
checkValidDFA(s);
return 0;
}
Python3
# Python3 program for the above approach
# Function to check whether the given
# is accepted by DFA or not
def checkValidDFA(s):
# Stores initial state of DFA
initial_state = 0
# Stores final state of DFA
final_state = 0
# Stores previous state of DFA
previous_state = 0
# Iterate through the string
for i in range(len(s)):
# Checking for all combinations
if ((s[i] == '0' and previous_state == 0) or
(s[i] == '1' and previous_state == 3)):
final_state = 1
elif ((s[i] == '0' and previous_state == 3) or
(s[i] == '1' and previous_state == 0)):
final_state = 2
elif ((s[i] == '0' and previous_state == 1) or
(s[i] == '1' and previous_state == 2)):
final_state = 0
elif ((s[i] == '0' and previous_state == 2) or
(s[i] == '1' and previous_state == 1)):
final_state = 3
# Update the previous_state
previous_state = final_state
# If final state is reached
if (final_state == 3):
print("Accepted")
# Otherwise
else:
print("Not Accepted")
# Driver Code
if __name__ == '__main__':
# Given string
s = "010011"
# Function Call
checkValidDFA(s)
# This code is contributed by mohit kumar 29
Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to check whether the given
// string is accepted by DFA or not
static void checkValidDFA(String s)
{
// Stores initial state of DFA
int initial_state = 0;
// Stores final state of DFA
int final_state = 0;
// Stores previous state of DFA
int previous_state = 0;
// Iterate through the string
for(int i = 0; i < s.length(); i++)
{
// Checking for all combinations
if ((s.charAt(i) == '0' && previous_state == 0) ||
(s.charAt(i) == '1' && previous_state == 3))
{
final_state = 1;
}
else if ((s.charAt(i) == '0' && previous_state == 3) ||
(s.charAt(i) == '1' && previous_state == 0))
{
final_state = 2;
}
else if ((s.charAt(i) == '0' && previous_state == 1) ||
(s.charAt(i) == '1' && previous_state == 2))
{
final_state = 0;
}
else if ((s.charAt(i) == '0' && previous_state == 2) ||
(s.charAt(i) == '1' && previous_state == 1))
{
final_state = 3;
}
// Update the previous_state
previous_state = final_state;
}
// If final state is reached
if (final_state == 3)
{
System.out.println("Accepted");
}
// Otherwise
else
{
System.out.println("Not Accepted");
}
}
// Driver Code
public static void main(String args[])
{
// Given string
String s = "010011";
// Function Call
checkValidDFA(s);
}
}
// This code is contributed by bgangwar59
C#
// C# program for the above approach
using System;
class GFG{
// Function to check whether the given
// string is accepted by DFA or not
static void checkValidDFA(string s)
{
// Stores initial state of DFA
//int initial_state = 0;
// Stores final state of DFA
int final_state = 0;
// Stores previous state of DFA
int previous_state = 0;
// Iterate through the string
for(int i = 0; i < s.Length; i++)
{
// Checking for all combinations
if ((s[i] == '0' && previous_state == 0) ||
(s[i] == '1' && previous_state == 3))
{
final_state = 1;
}
else if ((s[i] == '0' && previous_state == 3) ||
(s[i] == '1' && previous_state == 0))
{
final_state = 2;
}
else if ((s[i] == '0' && previous_state == 1) ||
(s[i] == '1' && previous_state == 2))
{
final_state = 0;
}
else if ((s[i] == '0' && previous_state == 2) ||
(s[i] == '1' && previous_state == 1))
{
final_state = 3;
}
// Update the previous_state
previous_state = final_state;
}
// If final state is reached
if (final_state == 3)
{
Console.WriteLine("Accepted");
}
// Otherwise
else
{
Console.WriteLine("Not Accepted");
}
}
// Driver Code
public static void Main()
{
// Given string
string s = "010011";
// Function Call
checkValidDFA(s);
}
}
// This code is contributed by sanjoy_62
Javascript
输出:
Accepted
时间复杂度: O(N)
辅助空间: O(1)
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